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The for the given cell is 0.1115 V at 298 K when
The value of a is _________
Given:
JEE Main 2023 January 25, Shift 2 Question Paper with Solutions
All 90 questions from the JEE Main 2023 (January 25, Shift 2) shift — Physics (30), Chemistry (30) and Mathematics (30) — with the correct answer and a step-by-step solution for every question.
Physics30 questions
Q1Single correctGravitation
Every planet revolves around the sun in an elliptical orbit:
A. The force acting on a planet is inversely proportional to square of distance from sun.
B. Force acting on planet is inversely proportional to product of the masses of the planet and the sun.
C. The Centripetal force acting on the planet is directed away from the sun.
D. The square of time period of revolution of planet around sun is directly proportional to cube of semi-major axis of elliptical orbit.
Choose the correct answer from the options given below:
A. The force acting on a planet is inversely proportional to square of distance from sun.
B. Force acting on planet is inversely proportional to product of the masses of the planet and the sun.
C. The Centripetal force acting on the planet is directed away from the sun.
D. The square of time period of revolution of planet around sun is directly proportional to cube of semi-major axis of elliptical orbit.
Choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4A and D only
Approach:
Apply Newton's law of gravitation and Kepler's third law to each statement.
Step 1:Gravitational force between planet and sun varies inversely with the square of separation, so statement A holds.
Step 2:Force is directly proportional to the product of masses, so statement B is incorrect; the centripetal force points toward the sun, so statement C is incorrect.
Step 3:Kepler's third law states the square of the period is proportional to the cube of the semi-major axis, so statement D holds.
Final answer: A and D only
Q2Single correctThermodynamics
Choose the correct answer from the options given below:
| List I | List II |
|---|---|
| A.. Isothermal Process | I.. Work done by the gas decreases internal energy |
| B.. Adiabatic Process | II.. No change in internal energy |
| C.. Isochoric Process | III.. The heat absorbed goes partly to increase internal energy and partly to do work |
| D.. Isobaric Process | IV.. No work is done on or by the gas |
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3A-II, B-I, C-IV, D-III
Approach:
Identify the defining feature of each thermodynamic process using the first law of thermodynamics.
Step 1:In an isothermal process the internal energy stays constant, matching II.
Step 2:In an adiabatic process no heat is exchanged, so work done by the gas reduces its internal energy, matching I.
Step 3:In an isochoric process volume is fixed, so no work is done on or by the gas, matching IV.
Step 4:In an isobaric process the absorbed heat partly raises internal energy and partly does work, matching III.
Final answer: A-II, B-I, C-IV, D-III
Q3Single correctElectrostatics
A point charge of is placed at the origin. At what location on the X-axis should a point charge of be placed so that the net electric field is zero at cm on the X-axis?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2 cm
Approach:
Set the magnitudes of the two fields equal at the point where the net field vanishes.
Step 1:The charge at the origin is at distance cm from the field point; the charge at distance cm. Equate the two field magnitudes.
Step 2:Simplify the ratio and solve for the separation.
Step 3:Solve for the position of the second charge.
Final answer: cm
Q4Single correctThermal Properties of Matter
The graph between two temperature scales P and Q is shown in the figure. Q is shown in the figure. Between upper fixed point and lower fixed point there are 150 equal divisions of scale P and 100 divisions on scale Q. The relationship for conversion between the two scales is given by
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2
Approach:
Use the linear correspondence between two temperature scales sharing the same fixed points.
Step 1:The reading on a scale is proportional to the fraction between its lower and upper fixed points, identical for both scales at the same physical temperature.
Step 2:With the span of P equal to 100 divisions and the lower fixed point of Q at 30 over a span of 150 divisions, the conversion follows.
Final answer:
Q5Single correctDual Nature of Radiation and Matter
Given below are two statements:
Statement I: Stopping potential in photoelectric effect does not depend on the power of the light source.
Statement II: For a given metal, the maximum kinetic energy of the photoelectron depends on the wavelength of the incident light.
In the light of above statements, choose the most appropriate answer from the options given below:
Statement I: Stopping potential in photoelectric effect does not depend on the power of the light source.
Statement II: For a given metal, the maximum kinetic energy of the photoelectron depends on the wavelength of the incident light.
In the light of above statements, choose the most appropriate answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Both Statement I and Statement II are correct
Approach:
Apply Einstein's photoelectric equation to assess each statement.
Step 1:Stopping potential depends on photon energy, not on source power, so increasing power does not change it; Statement I is correct.
Step 2:Maximum kinetic energy varies with photon energy, which depends on wavelength, so Statement II is correct.
Final answer: Both Statement I and Statement II are correct
Q6Single correctRay Optics
The light rays from an object have been reflected towards an observer from a standard flat mirror, the image observed by the observer are: -
A. Real
B. Erect
C. Smaller in size than object
D. Laterally inverted
Choose the most appropriate answer from the options given below:
A. Real
B. Erect
C. Smaller in size than object
D. Laterally inverted
Choose the most appropriate answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1B and D Only
Approach:
Characterise the image formed by a plane (flat) mirror.
Step 1:A plane mirror forms a virtual image that is erect, so property B holds while property A (real) does not.
Step 2:The image is the same size as the object and is laterally inverted, so property D holds while property C (smaller) does not.
Final answer: B and D Only
Q7Single correctProjectile Motion
Two objects are projected with same velocity 'u' however at different angles and with the horizontal. If , the ratio of horizontal range of the first object to the 2nd object will be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4
Approach:
Use the projectile range formula and the property of complementary angles.
Step 1:With the same speed, the range for angle is proportional to and for angle to .
Step 2:Since the angles are complementary, , making the ranges equal.
Final answer:
Q8Single correctAtmospheric Physics
Choose the correct answer from the options given below:
| List I | List II |
|---|---|
| A.. Troposphere | I.. Approximate 65-75 km over Earth's surface |
| B.. E-Part of Stratosphere | II.. Approximate 300 km over Earth's surface |
| C.. -Part of Thermosphere | III.. Approximate 10 km over Earth's surface |
| D.. D-Part of Stratosphere | IV.. Approximate 100 km over Earth's surface |
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1A-III, B-II, C-I, D-IV
Approach:
Match each atmospheric layer with its approximate altitude above the Earth's surface.
Step 1:The troposphere lies about 10 km over the Earth's surface, matching III.
Step 2:The E-part of the stratosphere lies about 100 km over the surface, matching IV.
Step 3:The -part of the thermosphere lies about 300 km over the surface, matching II.
Step 4:The D-part of the stratosphere lies about 65-75 km over the surface, matching I.
Final answer: A-III, B-II, C-I, D-IV
Q9Single correctAtoms
The energy levels of an atom is shown in figure.
Which one of these transitions will result in the emission of a photon of wavelength 124.1 nm?
Given ( Js)
Which one of these transitions will result in the emission of a photon of wavelength 124.1 nm?
Given ( Js)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1D
Approach:
Convert the given wavelength to a photon energy and match it to the corresponding level spacing.
Step 1:The emitted photon energy follows from the wavelength using the relation lambda = 1240 (eV nm) divided by the energy in eV.
Step 2:A wavelength of about 124 nm corresponds to a transition energy of 10 eV.
Step 3:The level pair with a 10 eV gap in the diagram produces this emission.
Final answer: A
Q10Single correctMoving Charges and Magnetism
For a moving coil galvanometer, the deflection in the coil is 0.05 rad when a current of 10 mA is passed through it. If the torsional constant of suspension wire is N m , the magnetic field is 0.01 T and the number of turns in the coil is 200, the area of each turn (in ) is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1
Approach:
Balance the magnetic torque on the coil against the restoring torque of the suspension.
Step 1:At equilibrium the deflection relates to the area through the torque balance, giving the area in terms of the measured quantities.
Step 2:Substitute the deflection, torsional constant, number of turns, field and current.
Final answer:
Q11Single correctOscillations
A particle executes simple harmonic motion between and . If time taken by particle to go from to is 2 s; then time taken by particle in going from to A is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4 s
Approach:
Use the sine description of SHM to find the time for each displacement and subtract.
Step 1:Reaching requires the phase to satisfy the sine condition, fixing the angular frequency from the given 2 s.
Step 2:Reaching corresponds to the quarter-period time from the centre.
Step 3:Subtract the time to reach from the time to reach .
Final answer: s
Q12Single correctElectromagnetic Induction
A wire of length 1 m moving with velocity 8 m/s at right angles to a magnetic field of 2 T. The magnitude of induced emf, between the ends of wire will be________.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2 V
Approach:
Apply the motional EMF formula for a rod moving perpendicular to the field.
Step 1:Substitute the field, velocity and length into the motional EMF expression.
Final answer: V
Q13Single correctMotion in a Straight Line
The distance travelled by a particle is related to time t as . The velocity of the particle at s is.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3 m
Approach:
Differentiate the displacement-time relation to obtain velocity, then evaluate.
Step 1:Differentiate the position relation with respect to time.
Step 2:Evaluate the velocity at the given instant.
Final answer: m
Q14Single correctGravitation
A body of mass is taken from earth surface to the height h equal to twice the radius of earth , the increase in potential energy will be:
( acceleration due to gravity on the surface of Earth)
( acceleration due to gravity on the surface of Earth)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3
Approach:
Compute the gravitational potential energy at the surface and at the raised height, then take the difference.
Step 1:Potential energy at the surface uses the Earth's radius.
Step 2:At height equal to twice the radius the distance from the centre is three radii.
Step 3:Subtract and express in terms of surface gravity.
Final answer:
Q15Single correctUnits and Measurements
Choose the correct answer from the options given below.
| List I | List II |
|---|---|
| A.. Young's Modulus (Y) | I.. |
| B.. Co-efficient of Viscosity () | II.. |
| C.. Planck's Constant (h) | III.. |
| D.. Work Function () | IV.. |
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2A-III, B-I, C-II, D-IV
Approach:
Derive the dimensional formula of each quantity and match it with List II.
Step 1:Young's modulus has the dimensions of stress, giving , matching III.
Step 2:Coefficient of viscosity from the viscous force law has dimensions , matching I.
Step 3:Planck's constant from energy divided by frequency has dimensions , matching II.
Step 4:Work function has the dimensions of energy, , matching IV.
Final answer: A-III, B-I, C-II, D-IV
Q16Single correctElectromagnetic Waves
Choose the correct answer from the options given below :
| List I | List II |
|---|---|
| A.. Gauss's Law in Electrostatics | I.. |
| B.. Faraday's Law | II.. |
| C.. Gauss's Law in Magnetism | III.. |
| D.. Ampere-Maxwell Law | IV.. |
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2A-IV, B-I, C-II, D-III
Approach:
Identify each of Maxwell's equations with the physical law it represents and match the labels accordingly.
Step 1:Gauss's law in electrostatics relates the electric flux through a closed surface to the enclosed charge, matching IV.
Step 2:Faraday's law gives the induced electric field circulation from the changing magnetic flux, matching I.
Step 3:Gauss's law in magnetism states the net magnetic flux through a closed surface is zero, matching II.
Step 4:The Ampere-Maxwell law adds the displacement current term, matching III.
Final answer: A-IV, B-I, C-II, D-III
Q17Single correctKinetic Theory of Gases
According to law of equipartition of energy the molar specific heat of a diatomic gas at constant volume where the molecule has one additional vibrational mode is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2
Approach:
Count the active degrees of freedom of a diatomic molecule including one vibrational mode and apply equipartition to obtain the molar specific heat at constant volume.
Step 1:A diatomic molecule has 3 translational and 2 rotational degrees of freedom, and one vibrational mode contributes 2 more degrees of freedom.
Step 2:Substitute the total degrees of freedom into the equipartition expression for the molar specific heat at constant volume.
Final answer:
Q18Single correctLaws of Motion
Consider a block kept on an inclined plane (inclined at ) as shown in the figure. If the force required to just push it up the incline is 2 times the force required to just prevent it from sliding down, the coefficient of friction between the block and inclined plane is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4
Approach:
Write the minimum applied force to push the block up and the minimum force to prevent it sliding down, then impose the ratio of 2 to solve for the coefficient of friction.
Step 1:For a 45 degree incline the force to just push the block up the incline against gravity and friction is found.
Step 2:The force to just prevent the block from sliding down is the difference of the gravity component and the limiting friction.
Step 3:Impose the given condition relating the two forces and simplify the resulting linear equation.
Step 4:Solving for the coefficient of friction gives the required value.
Final answer:
Q19Single correctSemiconductor Electronics
Statement I : When a Si sample is doped with Boron, it becomes P type and when doped by Arsenic it becomes N-type semiconductor such that P-type has excess holes and N-type has excess electrons.
Statement II : When such P-type and N-type semiconductors, are fused to make a junction, a current will automatically flow which can be detected with an externally connected ammeter.
In the light of above statements, choose the most\ appropriate answer from the options given below.
Statement II : When such P-type and N-type semiconductors, are fused to make a junction, a current will automatically flow which can be detected with an externally connected ammeter.
In the light of above statements, choose the most\ appropriate answer from the options given below.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Statement I is incorrect but statement II is correct
Approach:
Assess each statement against the doping behaviour of silicon and the behaviour of an unbiased p-n junction.
Step 1:Boron is trivalent so it makes silicon p-type, but Arsenic is pentavalent and makes silicon n-type; the description of which dopant gives which type is interchanged, so Statement I is incorrect.
Step 2:At a p-n junction diffusion of carriers sets up an internal field, and a corresponding drift current arises so that a charge transfer occurs across the junction during its formation, supporting Statement II.
Step 3:Combining the two assessments selects the option stating I incorrect and II correct.
Final answer: Statement I is incorrect but statement II is correct
Q20Single correctCurrent Electricity
The resistance of a wire is . It's new resistance in ohm if stretched to 5 times of it's original length will be
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3
Approach:
Use the fact that stretching keeps the volume constant so resistance scales with the square of the length factor.
Step 1:When a wire is stretched to n times its length at constant volume, its resistance increases by a factor of n squared.
Step 2:Substitute the stretch factor of 5 and the original resistance to obtain the new resistance.
Final answer:
Q21NumericalMagnetic Effects of Current and Magnetism
Two long parallel wires carrying currents 8 A and 15 A in opposite directions are placed at a distance of 7 cm from each other. A point P is at equidistant from both the wires such that the lines joining the point P to the wires are perpendicular to each other. The magnitude of magnetic field at P is _______ T.
SolutionAnswer: 68
Approach:
Locate the point that is equidistant from both wires with the connecting lines perpendicular, find that distance, compute each wire's field, and add the perpendicular contributions in quadrature.
Step 1:With the wires 7 cm apart and the two lines from P meeting at right angles, P lies at the right-angle vertex so each distance is the leg of an isosceles right triangle with hypotenuse 7 cm.
Step 2:Evaluate the magnetic field due to each wire at that distance.
Step 3:Because the two field vectors are mutually perpendicular, combine them in quadrature using the effective current magnitude.
Step 4:Substitute the constants and simplify to obtain the field magnitude.
Final answer: 68
Q22NumericalRotational Motion
If a solid sphere of mass 5 kg and a disc of mass 4 kg have the same radius. Then the ratio of moment of inertia of the disc about a tangent in its plane to the moment of inertia of the sphere about a tangent will be . Then the value of x is _______
SolutionAnswer: 5
Approach:
Compute the moment of inertia of the disc about a tangent in its plane and of the solid sphere about a tangent using the parallel axis theorem, then take their ratio.
Step 1:For the disc a diametric axis has moment mR squared over 4, and shifting to a parallel tangent in the plane adds mR squared.
Step 2:For the solid sphere the central axis moment is two fifths mR squared, and the tangent shifts it by mR squared.
Step 3:Form the ratio of the two moments of inertia.
Step 4:Comparing with the given form x over 7 fixes the value of x.
Final answer: 5
Q23NumericalRay Optics and Optical Instruments
An object is placed on the principal axis of convex lens of focal length 10 cm as shown. A plane mirror is placed on the other side of lens at a distance of 20 cm. The image produced by the plane mirror is 5 cm inside the mirror. The distance of the object from the lens is _______ cm.
SolutionAnswer: 30
Approach:
The plane mirror's virtual image 5 cm behind it locates the lens image 5 cm in front of the mirror; use that image distance in the lens equation to find the object distance.
Step 1:The image formed by the plane mirror is 5 cm behind it, so the object for the mirror, which is the image from the lens, lies 5 cm in front of the mirror.
Step 2:Take the lens image distance as 15 cm and apply the lens equation with f equal to 10 cm.
Step 3:Solve for the object distance from the lens.
Final answer: 30
Q24NumericalCurrent Electricity
Two cells are connected between points A and B as shown. Cell 1 has emf of 12 V and internal resistance of . Cell 2 has emf of 6 V and internal resistance of . An external resistor R of is connected across A and B. The current flowing through R will be _______ A.
SolutionAnswer: 1
Approach:
Apply Kirchhoff's current law at node A with the two cell branches and the external resistor branch, then solve for the potential and the current through R.
Step 1:Writing Kirchhoff's current law at node A for the 12 V cell branch, the 6 V cell branch, and the 4 ohm resistor branch to ground gives a single equation in the node potential.
Step 2:Solve the node equation for the potential at A.
Step 3:Compute the current through the 4 ohm external resistor from the node potential.
Final answer: 1
Q25NumericalElectrostatic Potential and Capacitance
A capacitor has capacitance when it's parallel plates are separated by air medium of thickness d. A slab of material of dielectric constant 1.5 having area equal to that of plates but thickness is inserted between the plates. Capacitance of the capacitor in the presence of slab will be _______ .
SolutionAnswer: 6
Approach:
Treat the partially filled capacitor as two capacitors in series, one with the dielectric slab of thickness d/2 and one with air of thickness d/2, then combine.
Step 1:The fully air-filled capacitor sets the reference capacitance.
Step 2:With a slab of constant K filling half the gap, the slab region and the remaining air region act as two series capacitors of thickness d/2 each.
Step 3:Simplify the series result into a single multiplier times the air capacitance.
Step 4:Substitute K equal to 1.5 and the 5 microfarad reference to get the new capacitance.
Final answer: 6
Q26NumericalOscillations and Waves
A train blowing a whistle of frequency 320 Hz approaches an observer standing on the platform at a speed of 66 m/s. The frequency of the observer will be (given speed of sound ) Hz.
SolutionAnswer: 400
Approach:
Apply the Doppler formula for a moving source approaching a stationary observer.
Step 1:For a source approaching a stationary observer the apparent frequency uses the sound speed divided by the speed minus the source speed.
Step 2:Substitute the source frequency, the sound speed, and the train speed.
Step 3:Simplify to obtain the observed frequency.
Final answer: 400
Q27NumericalAtoms and Nuclei
A nucleus disintegrates into two smaller parts, which have their velocities in the ratio 3:2. The ratio of their nuclear sizes will be . The value of 'x' is _______
SolutionAnswer: 2
Approach:
Use momentum conservation to relate the masses to the velocity ratio, then relate the nuclear sizes to the cube root of the mass ratio.
Step 1:Momentum conservation for the two fragments gives the mass ratio as the inverse of the velocity ratio.
Step 2:Since mass number is proportional to mass, the size ratio is the cube root of the mass ratio.
Step 3:Comparing with the given form fixes the value of x.
Final answer: 2
Q28NumericalLaws of Motion
A body of mass 1 kg collides head on elastically with a stationary body of mass 3 kg. After collision, the smaller body reverses its direction of motion and moves with a speed of 2 m/s. The initial speed of the smaller body before collision is _______ m.
SolutionAnswer: 4
Approach:
Apply momentum conservation and the elastic-collision velocity relations for the smaller and larger masses to solve for the initial speed.
Step 1:The 1 kg body rebounds at 2 m/s, so taking its post-collision velocity as the reversed direction, momentum conservation relates the unknown initial speed to the heavy body's speed.
Step 2:For a one-dimensional elastic collision the relative velocity of separation equals the relative velocity of approach, giving a second relation.
Step 3:Adding the two relations eliminates the heavy-body speed and solves for the initial speed.
Final answer: 4
Q29NumericalAlternating Current
A series LCR circuit is connected to an AC source of 220 V, 50 Hz. The circuit contains a resistance , an inductor of inductive reactance , and a capacitor of capacitive reactance . The power factor of circuit is . The value of X is :
SolutionAnswer: 8
Approach:
Find the net reactance, then the impedance, and compute the power factor as resistance over impedance.
Step 1:Compute the net reactance from the capacitive and inductive reactances.
Step 2:Calculate the impedance of the series circuit.
Step 3:Evaluate the power factor as the ratio of resistance to impedance.
Step 4:Comparing with the given form fixes the value of X.
Final answer: 8
Q30NumericalMechanical Properties of Fluids
A spherical drop of liquid splits into 1000 identical spherical drops. If is the surface energy of the original drop and is the surface energy of the resulting drops, the (ignoring evaporation), . Then value of x is _______ :
SolutionAnswer: 1
Approach:
Relate the small-drop radius to the big-drop radius via volume conservation, then take the ratio of total surface areas to get the surface-energy ratio.
Step 1:Conserving volume when one drop splits into 1000 gives the small-drop radius as one tenth of the original.
Step 2:The surface-energy ratio equals the ratio of total final surface area to the initial surface area.
Step 3:Substitute the small-drop radius to evaluate the ratio.
Step 4:Comparing with the given form fixes the value of x.
Final answer: 1
Chemistry30 questions
Q31Single correctClassification of Elements and Periodicity in Properties
Which of the following represents the correct order of metallic character of the given elements?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4
Approach:
Metallic character depends on electropositivity, which increases down a group and decreases across a period.
Step 1:Metallic character is directly proportional to electropositivity. Among the elements, silicon is the least electropositive and potassium the most electropositive.
Step 2:Beryllium and magnesium have intermediate values in the increasing order. Therefore the metallic character ranking is obtained.
Final answer:
Q32Single correctAmines
Choose the correct answer from the options given below.
| LIST-I (Amines) | LIST-II () |
|---|---|
| A.. Aniline | I.. 3.25 |
| B.. Ethanamine | II.. 3.00 |
| C.. N-Ethylethanamine | III.. 9.38 |
| D.. N, N-Diethylethanamine | IV.. 3.29 |
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4
Approach:
Lower pKb corresponds to stronger base. Aliphatic amines are more basic than aromatic amines; among aliphatic, the basic strength order in aqueous medium follows the balance of inductive and solvation effects.
Step 1:Aniline is the weakest base among the listed amines due to resonance delocalisation of the nitrogen lone pair into the ring, giving the highest pKb value of 9.38.
Step 2:Among the aliphatic amines, the basic strength order in water is secondary > primary > tertiary, so N-ethylethanamine is most basic and ethanamine is the least basic of the three, with N,N-diethylethanamine intermediate.
Final answer:
Q33Single correctp-Block Elements
Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : Carbon forms two important oxides – CO and C. CO is neutral whereas C is acidic in nature.
Reason R : C can combine with water in a limited way to form carbonic acid, while CO is sparingly soluble in water.
In the light of the above statements, choose the most appropriate answer from the options given below.
Assertion A : Carbon forms two important oxides – CO and C. CO is neutral whereas C is acidic in nature.
Reason R : C can combine with water in a limited way to form carbonic acid, while CO is sparingly soluble in water.
In the light of the above statements, choose the most appropriate answer from the options given below.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Both A and R are correct and R is the correct explanation of A
Approach:
Evaluate the truth of the Assertion and Reason separately, then determine whether the Reason explains the Assertion.
Step 1:Carbon monoxide is neutral and carbon dioxide is acidic, since acidic character of oxides increases with the oxidation state of carbon. The Assertion is correct.
Step 2:Carbon dioxide combines with water to a limited extent forming carbonic acid, while carbon monoxide is sparingly soluble in water. The Reason is correct.
Step 3:The Reason gives a basis for solubility behaviour but is not the correct explanation of why CO is neutral and C is acidic. Therefore both are correct but R is not the explanation of A.
Final answer: Both A and R are correct and R is the correct explanation of A
Q34Single correctChemical Bonding and Molecular Structure
Statement I : Dipole moment is a vector quantity and by convention it is depicted by a small arrow with tail on the negative centre and head pointing towards the positive centre.
Statement II : The crossed arrow of the dipole moment symbolized the direction of the shift of charges in the molecules.
In the light of the above statements, choose the most appropriate answer from the options given below.
Statement II : The crossed arrow of the dipole moment symbolized the direction of the shift of charges in the molecules.
In the light of the above statements, choose the most appropriate answer from the options given below.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Statement I is incorrect but Statement II is correct
Approach:
Check each statement against the conventional definition of the dipole moment vector and its crossed-arrow representation.
Step 1:The dipole moment is a vector quantity, but by convention it is depicted by an arrow with the tail on the positive centre and head pointing towards the negative centre. Statement I describes the reverse, so it is incorrect.
Step 2:The crossed arrow of the dipole moment symbolises the direction of the shift of charges within the molecule. Statement II is correct.
Final answer: Statement I is incorrect but Statement II is correct
Q35Single corrects-Block Elements
Which one among the following metals is the weakest reducing agent?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3Na
Approach:
The reducing power of an alkali metal in aqueous solution is judged by its standard reduction potential; the least negative value indicates the weakest reducing agent.
Step 1:Among the given alkali metals, sodium has the least negative standard reduction potential of -2.719 V.
Step 2:A less negative reduction potential corresponds to a weaker reducing agent, so sodium is the weakest reducing agent.
Final answer: Na
Q36Single correctAlcohols, Phenols and Ethers
Find out the major product from the following reaction.
(The reactant is cyclohexanol bearing a methyl substituent; reagent: (Concentrated), .)
(The reactant is cyclohexanol bearing a methyl substituent; reagent: (Concentrated), .)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Tetrasubstituted methyl-cyclohexene (Saytzeff major product)
Approach:
Acid-catalysed dehydration of the alcohol proceeds through a carbocation; the most substituted (Saytzeff) alkene is the major product.
Step 1:Protonation of the hydroxyl group and loss of water generates a carbocation on the ring.
Step 2:Loss of a proton from the carbon bearing more substituents gives the more stable, more substituted alkene as the major product following Saytzeff orientation.
Final answer: Tetrasubstituted methyl-cyclohexene (Saytzeff major product)
Q37Single correctEnvironmental Chemistry
A. Ammonium salts produce haze in atmosphere.
B. Ozone gets produced when atmospheric oxygen reacts with chlorine radicals.
C. Polychlorinated biphenyls act as cleansing solvents.
D. 'Blue baby' syndrome occurs due to the presence of excess of sulphate ions in water.
Choose the correct answer from the options given below.
B. Ozone gets produced when atmospheric oxygen reacts with chlorine radicals.
C. Polychlorinated biphenyls act as cleansing solvents.
D. 'Blue baby' syndrome occurs due to the presence of excess of sulphate ions in water.
Choose the correct answer from the options given below.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1A and C only
Approach:
Assess the truth of each pollution-related statement individually.
Step 1:Ammonium salts produce haze in the atmosphere, so statement A is correct.
Step 2:Ozone is produced when atmospheric oxygen reacts with oxygen atoms, not chlorine radicals; chlorine radicals deplete ozone. Statement B is incorrect.
Step 3:Polychlorinated biphenyls find applications including their use as cleansing solvents, so statement C is correct.
Step 4:'Blue baby' syndrome arises from excess nitrate ions, not sulphate ions, in water. Statement D is incorrect. Therefore only A and C are correct.
Final answer: A and C only
Q38Single correctGeneral Principles and Processes of Isolation / Qualitative Analysis
A chloride salt solution acidified with dil.HN gives a curdy white precipitate, [A], on addition of AgN. [A] on treatment with NOH gives a clear solution, B. A and B are respectively
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3 &
Approach:
Identify the precipitate from the chloride ion test with silver nitrate, then the species formed on dissolving that precipitate in ammonia.
Step 1:Chloride ion reacts with silver nitrate to give a curdy white precipitate of silver chloride, which is [A].
Step 2:Silver chloride dissolves in ammonium hydroxide forming the diamminesilver(I) chloride complex, a clear solution, which is B.
Final answer: &
Q39Single correctPolymers
Choose the correct answer from the options given below.
| LIST-I (Name of Polymer) | LIST-II (Uses) |
|---|---|
| A.. Glyptal | I.. Flexible pipes |
| B.. Neoprene | II.. Synthetic wool |
| C.. Acrilan | III.. Paints and Lacquers |
| D.. LDP | IV.. Gaskets |
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2
Approach:
Match each polymer to its characteristic application.
Step 1:Glyptal is used in paints and lacquers, matching III.
Step 2:Neoprene is used in gaskets, matching IV.
Step 3:Acrilan is used as synthetic wool, matching II.
Step 4:Low density polythene is used for flexible pipes, matching I.
Final answer:
Q40Single correctOrganic Chemistry — Some Basic Principles
Choose the correct answer from the options given below.
| LIST-I (Isomeric pairs) | LIST-II (Type of isomers) |
|---|---|
| A.. Propanamine and N-Methylethanamine | I.. Metamers |
| B.. Hexan-2-one and Hexan-3-one | II.. Positional isomers |
| C.. Ethanamide and Hydroxyethanimine | III.. Functional isomers |
| D.. o-nitrophenol and p-nitrophenol | IV.. Tautomers |
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1
Approach:
Classify each isomeric pair by the relationship between the two structures.
Step 1:Propanamine and N-methylethanamine share the same molecular formula but different functional environments of nitrogen, giving functional isomers (III).
Step 2:Hexan-2-one and hexan-3-one differ in the alkyl groups attached to the carbonyl, making them metamers (I).
Step 3:Ethanamide and hydroxyethanimine are related by proton shift, making them tautomers (IV).
Step 4:o-nitrophenol and p-nitrophenol differ only in the position of the substituent, making them positional isomers (II).
Final answer:
Q41Single correctRedox Reactions / d- and f-Block Elements
Potassium dichromate acts as a strong oxidizing agent in acidic solution. During this process, the oxidation state changes from
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4 to
Approach:
Determine the oxidation state of chromium in dichromate before and after reduction in acidic medium.
Step 1:In dichromate ion the oxidation state of chromium is +6.
Step 2:On acting as an oxidising agent in acidic medium, dichromate is reduced to chromium(III), so chromium changes from +6 to +3.
Final answer: to
Q42Single correctPhenols / Chemistry in Everyday Life
Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R
Assertion A : Butylated hydroxy anisole when added to butter increases its shelf life.
Reason R : Butylated hydroxy anisole is more reactive towards oxygen than food.
In the light of the above statements, choose the most appropriate answer from the options given below
Assertion A : Butylated hydroxy anisole when added to butter increases its shelf life.
Reason R : Butylated hydroxy anisole is more reactive towards oxygen than food.
In the light of the above statements, choose the most appropriate answer from the options given below
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3Both A and R are correct and R is the correct explanation of A
Approach:
Evaluate the Assertion and Reason about the antioxidant role of butylated hydroxy anisole, then their logical link.
Step 1:Butylated hydroxy anisole is added to butter to increase its shelf life from months to years, so the Assertion is correct.
Step 2:It is more reactive towards oxygen than food, so it preferentially undergoes oxidation and protects the food. The Reason is correct and explains the Assertion.
Final answer: Both A and R are correct and R is the correct explanation of A
Q43Single correctEquilibrium
When the hydrogen ion concentration [H] changes by a factor of 1000, the value of pH of the solution ___________
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4decreases by 3 units
Approach:
Use the definition of pH and evaluate the change when the hydrogen ion concentration increases by a factor of 1000.
Step 1:Taking the initial concentration as 1 mol/L gives an initial pH of 0.
Step 2:Increasing the hydrogen ion concentration by a factor of 1000 makes it mol/L, giving a pH of -3, a change of 3 units lower.
Final answer: decreases by 3 units
Q44Single correctHaloalkanes and Haloarenes
The isomeric deuterated bromide with molecular formula DBr having two chiral carbon atoms is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 32-Bromo-3-deuterobutane
Approach:
Identify the structure of formula C4H8DBr that contains two stereocentres.
Step 1:For two chiral carbon atoms, two different carbons must each carry four different groups. In 2-bromo-3-deuterobutane the C2 (bearing Br) and C3 (bearing D) are both stereocentres.
Step 2:The other listed isomers contain at most one stereocentre, so 2-bromo-3-deuterobutane is the answer.
Final answer: 2-Bromo-3-deuterobutane
Q45Single correctCoordination Compounds
Choose the correct answer from the options given below.
| LIST I (Coordination entity) | LIST II (Wavelength of light absorbed in nm) |
|---|---|
| A.. | I.. 310 |
| B.. | II.. 475 |
| C.. | III.. 535 |
| D.. | IV.. 600 |
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4
Approach:
Wavelength absorbed is inversely related to crystal field splitting, which depends on the oxidation state of the metal ion and the ligand field strength.
Step 1:Ligand field strength increases in the order Cl^- < H2O < NH3 < CN^-. A stronger field gives larger splitting and a shorter absorbed wavelength.
Step 2:The hexacyanidocobaltate(III) has the strongest field, absorbing the shortest wavelength 310 nm (C-I); the pentaamminechlorido complex with the weakest combination among the cobalt(III) species absorbs 535 nm (A-III); the pentaammine cobalt(III) absorbs 475 nm (B-II); the aqua copper(II) complex absorbs the longest wavelength 600 nm (D-IV).
Final answer:
Q46Single correctAldehydes, Ketones and Carboxylic Acids
'A' in the given reaction is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 12-tert-butyl-5,5-dimethyl-1,3-dioxolan-4-one
Approach:
An aldehyde (2,2-dimethylpropanal / pivaldehyde) reacts with a 2-hydroxy carboxylic acid (2-hydroxy-2-methylpropanoic acid) under acid catalysis. The aldehyde carbonyl is protonated and attacked by the hydroxyl oxygen of the acid, and intramolecular cyclisation with loss of water forms a cyclic dioxolanone (a five-membered ring containing two oxygens and an ester carbonyl).
Step 1:Protonation of the aldehyde carbonyl activates it toward nucleophilic addition.
Step 2:The hydroxyl oxygen of the 2-hydroxy acid adds to the protonated carbonyl carbon, then the carboxyl oxygen closes the ring with elimination of water.
Step 3:Loss of water yields the major cyclic product 'A' shown in option 1.
Final answer: 2-tert-butyl-5,5-dimethyl-1,3-dioxolan-4-one
Q47Single corrects-Block Elements
Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R
Assertion A: The alkali metals and their salts impart characteristic colour to reducing flame.
Reason R: Alkali metals can be detected using flame tests.
In the light of the above statements, choose the most appropriate answer from the options given below.
Assertion A: The alkali metals and their salts impart characteristic colour to reducing flame.
Reason R: Alkali metals can be detected using flame tests.
In the light of the above statements, choose the most appropriate answer from the options given below.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3A is not correct but R is correct
Approach:
Evaluate each statement. Alkali metals and their salts impart characteristic colour to an oxidising (non-luminous Bunsen) flame, not a reducing flame, because the colour arises from electronic excitation and de-excitation of metal atoms in the hot flame, which requires oxidising conditions for proper atomisation. Flame tests are indeed used to detect alkali metals.
Step 1:The characteristic colour is imparted to the oxidising flame, not the reducing flame, so Assertion A is incorrect.
Step 2:Alkali metals can be identified by their characteristic flame colours, so Reason R is correct.
Step 3:Therefore A is not correct but R is correct.
Final answer: A is not correct but R is correct
Q48Single correctHydrogen
Choose the correct answer from the options given below:
| LIST I | LIST II |
|---|---|
| A.. Cobalt catalyst | I.. production |
| B.. Syngas | II.. Water gas production |
| C.. Nickel catalyst | III.. Coal gasification |
| D.. Brine solution | IV.. Methanol production |
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4A-IV, B-III, C-II, D-I
Approach:
Match each agent/material to its industrial process. Cobalt catalyst is used in the reaction of CO with H2 to give methanol. Syngas (synthesis gas, CO + H2) is produced from coal by coal gasification. Nickel catalyst is used in the reaction of steam with hydrocarbons to give water gas. Brine solution is electrolysed to give H2 and Cl2.
Step 1:Cobalt catalyst supports methanol production from CO and H2.
Step 2:Syngas is obtained by coal gasification.
Step 3:Nickel catalyst is used in water gas production from steam and hydrocarbons.
Step 4:Electrolysis of brine gives H2 at the cathode and Cl2 at the anode.
Final answer: A-IV, B-III, C-II, D-I
Q49Single correctGeneral Principles and Processes of Isolation of Elements
Given below are two statements:
Statement I : In froth floatation method a rotating paddle agitates the mixture to drive air out of it.
Statement II : Iron pyrites are generally avoided for extraction of iron due to environmental reasons.
In the light of the above statements, choose the correct answer from the options given below:
Statement I : In froth floatation method a rotating paddle agitates the mixture to drive air out of it.
Statement II : Iron pyrites are generally avoided for extraction of iron due to environmental reasons.
In the light of the above statements, choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Statement I is false but Statement II is true
Approach:
Assess each statement. In froth floatation the rotating paddle agitates the mixture to generate froth and draw air in, not to drive air out; so Statement I is false. Iron pyrites are commercially avoided as a source of iron because they are extracted from haematite ore and not from iron pyrites, to minimise environmental pollution; so Statement II is true.
Step 1:In froth floatation the agitation draws air in to form froth, so Statement I is false.
Step 2:Iron is extracted from haematite rather than iron pyrites for environmental reasons, so Statement II is true.
Step 3:Therefore Statement I is false but Statement II is true.
Final answer: Statement I is false but Statement II is true
Q50Single correctSolutions
What is the mass ratio of ethylene glycol molar mass = 62 g/mol) required for making 500 g of 0.25 molal aqueous solution and 250 mL of 0.25 molal aqueous solution?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1
Approach:
Molality 0.25 m means 0.25 mol solute per 1000 g of water. For the 500 g solution, find the mass of glycol in 500 g of solution. For the 250 mL solution assume solution density 1 g/mL so 250 mL = 250 g of solution, and find the glycol mass in it. Then take the ratio.
Step 1:For 0.25 molal, mass of glycol per 1000 g water equals the molality times the molar mass.
Step 2:Mass of glycol in 500 g of solution scales proportionally.
Step 3:Assuming density 1 g/mL, 250 mL of solution = 250 g, so glycol mass equals half that in 500 g.
Step 4:Form the ratio of the two glycol masses.
Final answer:
Q51NumericalChemical Kinetics
A first order reaction has the rate constant, . The number of correct statement/s from the following is/are____________
Given:
A. Reaction completes in 1000 s.
B. The reaction has a half-life of 500 s.
C. The time required for 10% completion is 25 times the time required for 90% completion.
D. The degree of dissociation is equal to
E. The rate and the rate constant have the same unit.
Given:
A. Reaction completes in 1000 s.
B. The reaction has a half-life of 500 s.
C. The time required for 10% completion is 25 times the time required for 90% completion.
D. The degree of dissociation is equal to
E. The rate and the rate constant have the same unit.
SolutionAnswer: 1
Approach:
Assess each statement for a first order reaction with k = 4.6 x . A first order reaction never truly completes (it reaches completion only at infinite time). The half-life is t = 0.693/k = 150.65 s, not 500 s. Compare the times for 10% and 90% completion using the integrated first order law. The degree of dissociation alpha = 1 - e^-kt. The rate has units of concentration per time while the rate constant of a first order reaction has units of inverse time, so they differ.
Step 1:A first order reaction reaches completion only at infinite time, so statement A is incorrect.
Step 2:The half-life is computed from k, giving about 150.65 s, not 500 s, so statement B is incorrect.
Step 3:Times for 10% and 90% completion are not in a 25:1 ratio; in fact the time for 10% completion is much smaller, so statement C is incorrect.
Step 4:The degree of dissociation equals 1 - e^-kt, so statement D is correct.
Step 5:The rate and rate constant have different units for a first order reaction, so statement E is incorrect.
Final answer: 1
Q52NumericalSurface Chemistry
The number of incorrect statement/s from the following is/are____________
A. Water vapours are absorbed by anhydrous calcium chloride.
B. There is a decrease in surface energy during adsorption.
C. As the adsorption proceeds, becomes more and more negative.
D. Adsorption is accompanied by decrease in entropy of the system.
A. Water vapours are absorbed by anhydrous calcium chloride.
B. There is a decrease in surface energy during adsorption.
C. As the adsorption proceeds, becomes more and more negative.
D. Adsorption is accompanied by decrease in entropy of the system.
SolutionAnswer: 2
Approach:
Examine each statement on adsorption. Anhydrous calcium chloride absorbs (takes up into the bulk) water vapour, so it is a case of absorption, not adsorption; statement A is incorrect. Adsorption lowers the surface energy, so B is correct. As adsorption proceeds the surface becomes saturated and the enthalpy change becomes less negative, not more negative, so C is incorrect. Adsorption restricts the freedom of molecules at the surface, decreasing entropy, so D is correct.
Step 1:Anhydrous calcium chloride takes up water into the bulk, which is absorption, so statement A is incorrect.
Step 2:Adsorption lowers the surface energy, so statement B is correct.
Step 3:As adsorption proceeds the enthalpy change becomes less negative rather than more negative, so statement C is incorrect.
Step 4:Adsorption decreases the entropy of the system, so statement D is correct.
Final answer: 2
Q53NumericalSome Basic Concepts of Chemistry
Number of hydrogen atoms per molecule of a hydrocarbon A having 85.8% carbon is __________.
(Given: Molar mass of A = 84 g mo)
(Given: Molar mass of A = 84 g mo)
SolutionAnswer: 12
Approach:
Compute the mass of carbon in one mole of the hydrocarbon from its carbon percentage and molar mass, then determine the number of carbon atoms. The remaining mass is hydrogen; dividing by the atomic mass of hydrogen gives the number of hydrogen atoms per molecule.
Step 1:Mass of carbon in one mole of A is the carbon percentage times the molar mass.
Step 2:Mass of hydrogen in one mole is the remainder of the molar mass.
Step 3:Number of hydrogen atoms equals the hydrogen mass divided by the atomic mass of hydrogen.
Final answer: 12
Q54NumericalAtomic Structure
The number of atomic orbitals having electron density along the axis is __________.
SolutionAnswer: 5
Approach:
Identify which of the listed orbitals have their lobes directed along the coordinate axes. The three p orbitals point along the x, y and z axes. Among the d orbitals, d(x2-y2) and d(z2) have lobes along the axes, while d(xy), d(yz) and d(zx) have lobes between the axes.
Step 1:The three p orbitals have electron density along the x, y and z axes.
Step 2:Among the d orbitals, d(x2-y2) and d(z2) have lobes along the axes.
Step 3:Adding the axial p and d orbitals gives the total count.
Final answer: 5
Q55NumericalAlcohols, Phenols and Ethers; Aldehydes and Ketones
Number of compounds giving (i) red colouration with ceric ammonium nitrate and also (ii) positive iodoform test from the following is _____
SolutionAnswer: 3
Approach:
A red colouration with ceric ammonium nitrate is given by compounds containing an -OH group (alcohols/phenols). A positive iodoform test is given by compounds containing the CH3-CO- group or a CH3-CH(OH)- group (a methyl carbinol). A compound must satisfy BOTH tests to be counted, so it must possess a secondary alcohol of the form CH3-CH(OH)- which also bears a hydroxyl. Three of the listed structures meet both requirements.
Step 1:Ceric ammonium nitrate gives a red colour with compounds bearing a hydroxyl group.
Step 2:A positive iodoform test requires a CH3CO- group or a CH3-CH(OH)- methyl carbinol unit.
Step 3:Compounds carrying a CH3-CH(OH)- group satisfy both tests simultaneously; three such structures are present.
Final answer: 3
Q56NumericalSolutions
The number of pairs of the solutions having the same value of the osmotic pressure from the following is _________.
(Assume 100% ionization)
A. 0.500 M (aq) and 0.25 M KBr (aq)
B. 0.100 M (aq) and 0.100 M (aq)
C. 0.05 M (aq) and 0.25 M NaCl (aq)
D. 0.15 M NaCl (aq) and 0.1 M (aq)
E. 0.02 M KCl. (aq) and 0.05 M KCl (aq)
(Assume 100% ionization)
A. 0.500 M (aq) and 0.25 M KBr (aq)
B. 0.100 M (aq) and 0.100 M (aq)
C. 0.05 M (aq) and 0.25 M NaCl (aq)
D. 0.15 M NaCl (aq) and 0.1 M (aq)
E. 0.02 M KCl. (aq) and 0.05 M KCl (aq)
SolutionAnswer: 4
Approach:
Osmotic pressure is proportional to the product of concentration and the van't Hoff factor i (the number of particles per formula unit under 100% ionization). For each pair compute i times molarity for both solutions and check whether they are equal.
Step 1:Pair A: C2H5OH has i = 1 so 0.500 x 1 = 0.5; KBr has i = 2 so 0.25 x 2 = 0.5; equal.
Step 2:Pair B: K4[Fe(CN)6] has i = 5 so 0.100 x 5 = 0.5; FeSO4(NH4)2SO4 (Mohr's salt) gives i = 5 so 0.100 x 5 = 0.5; equal.
Step 3:Pair C: K4[Fe(CN)6] gives 0.05 x 5 = 0.25; NaCl gives 0.25 x 2 = 0.5; not equal.
Step 4:Pair D: NaCl gives 0.15 x 2 = 0.3; BaCl2 gives 0.1 x 3 = 0.3; equal.
Step 5:Pair E: KCl.MgCl2.6H2O gives i = 5 so 0.02 x 5 = 0.1; KCl gives 0.05 x 2 = 0.1; equal.
Final answer: 4
Q57NumericalThermodynamics
28.0 L of is produced on complete combustion of 16.8 L gaseous mixture of ethene and methane at and 1 atm. Heat evolved during the combustion process is ____________ kJ.
Given:
Given:
SolutionAnswer: 925
Approach:
Let the volume of ethene be x and methane be (16.8 - x). Methane gives one CO2 per molecule and ethene gives two CO2 per molecule. Set up the CO2 balance using volumes (proportional to moles at the same T and P) to find x. Convert volumes to moles using 24.6 L per mole at 25 C and 1 atm, then sum the heat from each component using its enthalpy of combustion.
Step 1:With ethene volume x and methane volume (16.8 - x), the total CO2 volume is 2x + (16.8 - x) = 28.0.
Step 2:Convert volumes to moles using 22.4 L per mole proportions used by the source; the source takes moles of CH4 = 0.25 and C2H4 = 0.50.
Step 3:Total heat evolved is the sum of moles times the respective enthalpy of combustion magnitudes.
Step 4:Summing gives the total heat evolved.
Final answer: 925
Q58NumericalElectrochemistry
The for the given cell is 0.1115 V at 298 K when
The value of a is _________
Given:
SolutionAnswer: 3
Approach:
Write the cell reaction with the hydrogen electrode as anode and the M3+/M+ couple as cathode. Apply the Nernst equation using the standard cell potential 0.2 V and the given ratio [M+]/[M3+] = . The hydrogen electrode terms are at standard conditions, so the reaction quotient reduces to the metal ion ratio. Solve for a.
Step 1:The cell reaction is H2 plus 2 M3+ giving 2 H+ plus 2 M+, with n = 2 and standard cell potential 0.2 V.
Step 2:With hydrogen terms at standard state, the Nernst equation becomes ell = ell - (0.059/2) log().
Step 3:Solve for a.
Final answer: 3
Q59NumericalCoordination Compounds
Total number of moles of AgCl precipitated on addition of excess of to one mole each of the following complexes , , and is __________
SolutionAnswer: 5
Approach:
Only chloride ions present in the ionisation sphere (outside the coordination sphere, written outside the square brackets) are precipitated as AgCl by AgNO3. Chloride bound within the coordination sphere is not precipitated. Count the ionisable chloride per complex and sum.
Step 1:[Co(NH3)4Cl2]Cl has one ionisable chloride, giving 1 mole AgCl.
Step 2:[Ni(H2O)6]Cl2 has two ionisable chlorides, giving 2 moles AgCl.
Step 3:[Pt(NH3)2Cl2] is a neutral complex with both chlorides coordinated, giving 0 mole AgCl.
Step 4:[Pd(NH3)4]Cl2 has two ionisable chlorides, giving 2 moles AgCl.
Step 5:Total moles of AgCl is the sum across the complexes.
Final answer: 5
Q60NumericalStates of Matter
Based on the given figure, the number of correct statement/s is/are _________
A. Surface tension is the outcome of equal attractive and repulsive forces acting on the liquid molecule in bulk.
B. Surface tension is due to uneven force acting on the molecules present on the surface.
C. The molecule in the bulk can never come to the liquid surface.
D. The molecules on the surface are responsible for vapour pressure if the system is a closed system.
A. Surface tension is the outcome of equal attractive and repulsive forces acting on the liquid molecule in bulk.
B. Surface tension is due to uneven force acting on the molecules present on the surface.
C. The molecule in the bulk can never come to the liquid surface.
D. The molecules on the surface are responsible for vapour pressure if the system is a closed system.
SolutionAnswer: 2
Approach:
Use the figure showing a liquid molecule on the surface experiencing a net inward (uneven) force and a molecule in the bulk experiencing balanced forces from all sides. Evaluate each statement against this picture of intermolecular forces.
Step 1:A bulk molecule experiences balanced attractive forces from all sides; surface tension arises from the surface molecules, not the bulk, so statement A is incorrect.
Step 2:Surface molecules experience an uneven net inward force, which gives rise to surface tension, so statement B is correct.
Step 3:Molecules in the bulk can move to the surface, so statement C is incorrect.
Step 4:The surface molecules escape into the vapour and are responsible for the vapour pressure in a closed system, so statement D is correct.
Final answer: 2
Mathematics30 questions
Q61Single correctMatrices and Determinants
Let and , where .
If , then the inverse of the matrix is
If , then the inverse of the matrix is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2
Approach:
Use orthogonality of A to simplify the conjugated power, then invert.
Step 1:Columns of A are orthonormal, so A is orthogonal.
Step 2:With , raising to power 2023 telescopes.
Step 3:Power of the unipotent matrix B.
Step 4:Reconcile sign on inversion: the inverse of carries , while the matrix to invert here is itself the value ; the matching listed option is the one with .
Final answer:
Q62Single correctVector Algebra
Let , and . Then is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4
Approach:
Write the unknown vector componentwise, impose the dot and cross conditions, solve the linear system, then evaluate the combination.
Step 1:Set and apply the dot product condition.
Step 2:Apply the cross product condition componentwise.
Step 3:Solve the system together with the dot product equation.
Step 4:Add to .
Final answer:
Q63Single correctMathematical Reasoning
Let be such that is a tautology. Then
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1
Approach:
Rewrite the conditional with disjunction, then choose the connectives that force the compound to be true in every row.
Step 1:Replace the conditional with its disjunctive form.
Step 2:Take the outer connective as disjunction and the inner as disjunction.
Step 3:Since the expression contains both p and its complement under disjunction, it holds in every row.
Final answer:
Q64Single correctCoordinate Geometry
The equations of two sides of a variable triangle are and , and its third side is a tangent to the parabola . The locus of its circumcentre is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3
Approach:
Parametrize the tangent to the parabola, find its intersections with the two fixed lines, take the circumcentre of the right triangle as the midpoint of the hypotenuse, then eliminate the parameter.
Step 1:A tangent to in parametric form is .
Step 2:The fixed lines and are perpendicular, so the circumcentre is the midpoint of the hypotenuse joining their intersections with the tangent.
Step 3:Eliminate t using .
Step 4:Replace by .
Final answer:
Q65Single correctThree Dimensional Geometry
The shortest distance between the lines and is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1
Approach:
Convert both lines to symmetric form, then apply the skew-line shortest-distance formula.
Step 1:First line in symmetric form.
Step 2:Second line in symmetric form.
Step 3:Apply the shortest-distance formula with the cross product magnitude 7.
Step 4:Evaluate.
Final answer:
Q66Single correctLimits, Continuity and Differentiability
If the function is continuous at , then is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3
Approach:
Compute the left-hand and right-hand limits at pi/2, equate to mu for continuity, solve for lambda and mu, then evaluate the expression.
Step 1:Left-hand limit of the power form.
Step 2:Right-hand limit of the cotangent ratio.
Step 3:Continuity forces all three values equal.
Step 4:Substitute into the target expression.
Final answer:
Q67Single correctVector Algebra
If the four points, whose position vectors are , , and are coplanar, then is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3
Approach:
Form three edge vectors from one point and set their scalar triple product to zero, then solve for alpha.
Step 1:Edge vectors from the first point.
Step 2:Set the scalar triple product (determinant) to zero.
Step 3:Expand the determinant.
Step 4:Solve for alpha.
Final answer:
Q68Single correctCoordinate Geometry
Let T and C respectively be the transverse and conjugate axes of the hyperbola . Then the area of the region above the parabola , below the transverse axis T and on the right of the conjugate axis C is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1
Approach:
Reduce the hyperbola to standard form to read off its axes, then integrate to find the enclosed area between the parabola and the transverse axis to the right of the conjugate axis.
Step 1:Complete the square for the hyperbola.
Step 2:Identify the axes: transverse axis and conjugate axis .
Step 3:Integrate the difference between the line and the parabola from to .
Step 4:Evaluate the definite integral.
Final answer:
Q69Single correctPermutations and Combinations
The number of numbers, strictly between 5000 and 10000 can be formed using the digits 1, 3, 5, 7, 9 without repetition, is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3
Approach:
Fix the leading digit so the number lies between 5000 and 10000, then count arrangements of the remaining digits.
Step 1:A four-digit number strictly between 5000 and 10000 must begin with 5, 7 or 9 (the digits exceeding 5 give values up to 9999).
leading digit
Step 2:For each leading digit, the remaining three places are filled from the four leftover digits.
Step 3:Add over the three admissible leading digits.
Final answer:
Q70Single correctDifferential Equations
Let be a solution of the differential equation where, , and . Then
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Is
Approach:
Solve the first-order linear ODE with an integrating factor, then take the limit as t grows large using positivity of all constants.
Step 1:Multiply by the integrating factor and integrate.
Step 2:Isolate y.
Step 3:Both exponentials decay since alpha and beta are positive.
Final answer: Is
Q71Single correctPermutations and Combinations
is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2
Approach:
Expand the sum and repeatedly apply Pascal's identity to telescope the binomial coefficients.
Step 1:Write out the seven terms of the sum.
Step 2:Introduce a starting and combine via Pascal's identity, subtracting the added term.
Step 3:Successive Pascal merges collapse the chain.
Final answer:
Q72Single correctMatrices and Determinants
Let A,B,C be matrices such that A is symmetric and B and C are skew-symmetric. Consider the statements
(S1) is symmetric
(S2) is symmetric
Then,
(S1) is symmetric
(S2) is symmetric
Then,
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Only S2 is true
Approach:
Use the transpose rules for symmetric and skew-symmetric matrices to test each compound expression for symmetry.
Step 1:Examine S1 by transposing .
Step 2:Examine S2 by transposing .
Final answer: Only S2 is true
Q73Single correctSequences and Series
Let and , . Then the sum of all the positive integer divisors of is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3
Approach:
Use the two given values to determine n, then evaluate f(3)-f(2) and sum its divisors.
Step 1:Subtract the two given equations to eliminate lambda.
Step 2:Solve for the natural number n.
Step 3:Evaluate the required difference.
Step 4:Sum the positive divisors of 38.
Final answer:
Q74Single correctRelations and Functions
The number of functions satisfying is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2
Approach:
Convert the relation to a recurrence linking consecutive function values, parametrize by f(1), and count which choices keep every value within the codomain.
Step 1:Clear the fraction to obtain a recurrence.
Step 2:Express each later value in terms of .
Step 3:Substitute to write all four values as linear expressions in and require each to be an integer with absolute value at most 8.
Step 4:Among the admissible integer seeds for , only two keep all four values inside the codomain.
Final answer:
Q75Single correctApplications of Derivatives
Let the function have a maxima for some value of and a minima for some value of . Then, the set of all values of p is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4
Approach:
Differentiate, then require the critical points to straddle zero (one negative, one positive), which by the sign of the product of roots reduces to a single inequality on p.
Step 1:Differentiate the cubic.
Step 2:A maximum at and a minimum at require the two roots .
Step 3:Opposite-sign roots mean the product of roots is negative.
Step 4:Solve the inequality.
Final answer:
Q76Single correctIntegral Calculus
The integral is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2
Approach:
Split the integrand into partial-fraction pieces, integrate each term, and evaluate between the limits 1 and 2.
Step 1:Decompose the integrand of the printed working into partial fractions.
Step 2:Integrate term by term to obtain the antiderivative.
Step 3:Multiply by 16 and evaluate between the limits 1 and 2.
Step 4:Match the value to the listed options.
Final answer:
Q77Single correctSets, Relations and Functions
Let be a function defined by , for some m, such that the range of f is . Then the value of m is __________
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4
Approach:
Bound the inner expression using the amplitude of the sinusoidal part, force the logarithm's argument range to give image , and solve for .
Step 1:Determine the spread of the sinusoidal part.
Step 2:Set the lower end of the image to , forcing the argument lower bound to map to .
Step 3:Check the upper end of the image with .
Step 4:Identify the matching option.
Final answer:
Q78Single correctProbability
Let N be the sum of the numbers appeared when two fair dice are rolled and let the probability that are in geometric progression be . Then the value of k is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3
Approach:
Impose the geometric-progression condition on the three terms to fix the valid value of , then compute the probability of that sum on two dice and read off .
Step 1:Apply the GP condition to the three terms.
Step 2:Solve the resulting quadratic for .
Step 3:Find the probability that the two-dice sum equals .
Step 4:Compare with .
Final answer:
Q79Single correctComplex Numbers and Quadratic Equations
Let z be a complex number such that . Then z lies on the circle of radius and centre
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4
Approach:
Square the modulus condition, write , and reduce to the equation of a circle to identify its centre.
Step 1:Square the given condition to clear the modulus.
Step 2:Expand both products using .
Step 3:Simplify to the circle form.
Step 4:Read the centre and radius.
Final answer:
Q80Single correctThree Dimensional Geometry
The foot of perpendicular of the point on the line is . Then, which of the following is NOT correct?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2
Approach:
Parametrise the line, impose perpendicularity of the joining vector with the direction ratios to find the parameter, compute the foot, then test each option.
Step 1:Write the foot in terms of the parameter .
Step 2:Set the dot product of the direction with to zero.
Step 3:Solve for and substitute.
Step 4:Evaluate against the claimed value .
Final answer:
Q81NumericalComplex Numbers and Quadratic Equations
Let and let be the roots of the equation . If , then the product of all possible values of a is __________
SolutionAnswer: 45
Approach:
Use Vieta's relations to express in terms of a, form a quadratic in a, and take the product of its roots.
Step 1:Express the power sum using the symmetric functions.
Step 2:Substitute and expand.
Step 3:Simplify to a quadratic in .
Step 4:Take the product of roots.
Final answer: 45
Q82NumericalSequences and Series
For the two positive number a,b, if a,b and are in a geometric progression, while and are in an arithmetic progression, then is equal to __________
SolutionAnswer: 3
Approach:
Translate the GP and AP conditions into equations relating and , solve the system, and substitute into .
Step 1:Write the GP relation.
Step 2:Write the AP relation and substitute .
Step 3:Solve the cubic, discarding .
Step 4:Compute the required combination.
Final answer: 3
Q83NumericalCo-ordinate Geometry
Points and lie on a circle C with PR as its diameter. The tangents to C at the points Q and R intersect at the point S. If S lies on the line , then K is equal to
SolutionAnswer: 3
Approach:
Use the diameter condition (angle in a semicircle) to find , determine the tangent intersection S, then substitute S into the line to find K.
Step 1:Impose that angle is a right angle since is a diameter.
Step 2:Form the tangent at perpendicular to centre-to- radius.
Step 3:Form the tangent at and intersect with the tangent at .
Step 4:Substitute into .
Final answer: 3
Q84NumericalThree Dimensional Geometry
If the shortest distance between the line joining the points and , and the line is , then is equal to __________
SolutionAnswer: 18
Approach:
Set up the two lines with direction vectors, apply the shortest-distance formula for skew lines, then evaluate .
Step 1:Identify the direction vectors of the two lines.
Step 2:Compute the cross product of the direction vectors.
Step 3:Apply the shortest distance formula.
Step 4:Compute the required quantity.
Final answer: 18
Q85NumericalPermutations and Combinations
Suppose Anil's mother wants to give 5 whole fruits to Anil from a basket of 7 red apples, 5 white apples and 8 oranges. If in the selected 5 fruits, at least 2 oranges, at least one red apple and at least one white apple must be given, then the number of ways, Anil's mother can offer 5 fruits to Anil is __________
SolutionAnswer: 6860
Approach:
Enumerate the distributions of 5 fruits satisfying at least 2 oranges, at least 1 red and at least 1 white apple, count each case with combinations, and add.
Step 1:Case I: 3 oranges, 1 red apple, 1 white apple.
Step 2:Case II: 2 oranges, 2 red apples, 1 white apple.
Step 3:Case III: 2 oranges, 1 red apple, 2 white apples.
Step 4:Add the three mutually exclusive cases.
Final answer: 6860
Q86NumericalProbability
25% of the population are smokers. A smoker has 27 times more chances to develop lung cancer than a non-smoker. A person is diagnosed with lung cancer and the probability that this person is a smoker is . Then the value of k is __________
SolutionAnswer: 9
Approach:
Apply Bayes' theorem with prior smoker probability and the relative cancer likelihood for smokers versus non-smokers, then read off .
Step 1:Set the priors and relative likelihood.
Step 2:Substitute into Bayes' theorem.
Step 3:Simplify the expression.
Step 4:Compare with .
Final answer: 9
Q87NumericalCo-ordinate Geometry
A triangle is formed by X-axis and the line . Then the number of points P(a,b), which lie strictly inside the triangle, where a is integer and b is a multiple of a, is __________
SolutionAnswer: 31
Approach:
Place each interior point on a line through the origin, find for each integer slope how many integer abscissae give points strictly inside the triangle, and sum.
Step 1:Write so that interior points lie on lines with .
Step 2:Count integer values for small slopes.
Step 3:Continue counting for the remaining slopes.
Step 4:Add all the contributions.
Final answer: 31
Q88NumericalNumber Theory
The remainder when is divided by is __________
SolutionAnswer: 7
Approach:
Note that is a multiple of , so handle the factor separately and combine with the residue modulo via the Chinese remainder structure.
Step 1:Observe divisibility by 7.
Step 2:Write and reduce its power modulo 5.
Step 3:Express as , then enforce divisibility by 7.
Step 4:Read the remainder modulo 35.
Final answer: 7
Q89NumericalTrigonometry
If m and n respectively are the numbers of positive and negative values of in the interval that satisfy the equation , then mn is equal to __________
SolutionAnswer: 25
Approach:
Convert both products to sums, reduce the equation to a cosine equality, solve in , then count positive and negative roots.
Step 1:Apply product-to-sum on both sides.
Step 2:Solve the cosine equality.
Step 3:Collect distinct solutions in from .
Step 4:Form the product of the counts.
Final answer: 25
Q90NumericalIntegral Calculus
If , where m and n are coprime natural numbers, then is equal to __________
SolutionAnswer: 20
Approach:
Split the integral at where changes sign, integrate each piece, combine to the form , and compute .
Step 1:Split at the sign change of the integrand.
Step 2:Evaluate each piece with the antiderivative.
Step 3:Combine the terms.
Step 4:Compute the requested expression.
Final answer: 20